Compound 'A' with molecular formula `C_(4)H_(9)Br` is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound 'A' only. When another optically active isomer 'B' of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both.
(i) Write down the structural formula of both compounds 'A' and 'B'.
(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

Explanation

(i) The reaction procceeds through `S_(N)I` path as the given substrate is `3^(@)` - alkyl halide.
`underset("2-Bromo-2-methylpropane")(CH_(3)-overset(CH_(3))overset("| ")underset("Br ")underset("| ")("C ")-CH_(3))overset(KOH_((aq)))rarr underset("2-Methylpropan-2-ol")(CH_(3)-overset(CH_(3))overset("| ")underset("OH")underset("| ")("C ")-CH_(3))`
(ii) Compound (B) forms a product of inverted configuration as it undergoes `S_(N)2` reaction.